Correct option is b) It activates an activator protein.
Step-by-step explanation
- The lac operon is an inducible operon that utilizes lactose as an energy source and is activated when glucose is low and lactose is present.
- In bacteria such as E. coli, it appears that the products of carbohydrate metabolism themselves activate the switch between glucose and lactose use.
- Specifically, the catabolites produced by the breakdown of glucose prevent production of a signal molecule called cyclic AMP (cAMP).
- In turn, the absence of cAMP results in a failure to activate production of the enzymes necessary to metabolize lactose, even when lactose is available.
- Glucose therefore stops activation of the lac operon , which prevents lactose use and leads to preferential use of glucose.
- However, in the absence of glucose, the lac operon can be activated if the lac i repressor is turned off by lactose binding.
- In E. coli, the switch from glucose use to lactose use depends on the presence of both cAMP and CAP
- CAP binds with cAMP to form CAP-cAMP complex
- The CAP-cAMP complex then binds to a specific DNA sequence found upstream of the lac operon operator and promoter.
- CAP-cAMP complex binding leads to enhanced RNA polymerase binding and activation of gene expression from the lac operon.
- This process is affected by glucose levels, because cAMP levels are decreased in the presence of glucose catabolites.
- Thus, an elevation in cAMP concentration signals the absence of glucose, because lower glucose levels lead to increased cAMP levels.
- In turn, increased cAMP levels lead to enhanced expression of the lac operon.
- In the presence of glucose, however, intracellular levels of cAMP fall, which leads to a lack of lac operon activation.
- The lac operon is therefore positively regulated by the absence of glucose catabolites